Question

To what series does the spectral line of atomic hydrogen belong if its wave number is equal to the difference between the wave numbers of the following two lines of the Balmer series: 486.1nm and 410.2nm?
A. lyman series
B. balmer series
C. paschen series
D. bracket series

Answer 1

Hint: The hydrogen spectrum comprises different series which have different energies associated with them. The energy can be linked to the wavelength which was given by Rydberg and the wavelength decides the regions of the spectrum.

Complete answer:
-Hydrogen atom consists of 1 proton and 1 electron. The energy state of the orbitals of the atom was given by Bohr in his atomic model and the spectral emission is based on Schrodinger’s equation.
-Electrons jump from higher energy level to lower energy level releasing energy in form of spectral emissions. The spectral lines are grouped together to form the spectral series. Hydrogen shows 4 such series named Lyman, Balmer, Brackett and Paschen.
-The energy differences and the wavelengths of the emitted photons is related by Rydberg formula as
 $${\displaystyle \frac{1}{\lambda }}=Z^{2}R({\displaystyle \frac{1}{{n_1}^2}}-{\displaystyle \frac{1}{{n_2}^2}})$$
Where Z is the atomic number, n is the principal quantum number and the numbers 1 and 2 represent the lower and higher energy levels and R is the Rydberg’s constant.
-Different series are obtained by changing the values of n as 1,2,3 and 4 for hydrogen. The value of R for hydrogen is 1.09677x${10}^7{m}^{-1}$ .
-In the question we are given 2 different wavelengths and we can find the 2 different values of n from this using the formula to get the transition and then equate it again to get the wavelength and thus the series.
 $\begin{array}{rl}& \nu =[{\displaystyle \frac{1}{{\lambda }_2}}-{\displaystyle \frac{1}{{\lambda }_1}}]\\ & \Rightarrow \nu \text{=}R({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{{n_2}^2}})-R({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{{n_1}^2}})\\ & \Rightarrow \nu =R({\displaystyle \frac{1}{{n_1}^2}}-{\displaystyle \frac{1}{{n_2}^2}})\end{array}$
-Now for line 1 of Balmer, we can write
     $$\begin{array}{rl}& {\displaystyle \frac{1}{{\lambda }_1}}=R({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{{n_1}^2}})\\ & \Rightarrow {\displaystyle \frac{1}{481.6x{10}^{-7}}}=109678x({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{{n_1}^2}})\\ & \Rightarrow n_1=4\end{array}$$
-For line 2 of Balmer series,
     $$\begin{array}{rl}& {\displaystyle \frac{1}{{\lambda }_2}}=R({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{{n_2}^2}})\\ & \Rightarrow {\displaystyle \frac{1}{410.2x{10}^{-7}}}=109678x({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{{n_2}^2}})\\ & \Rightarrow n_2=6\end{array}$$

So the transition of electrons occurs from shell 6 to shell 4 and so the series is paschen series and the correct option is C.

Note:
The wavelength for the series can be found as
. $$\begin{array}{rl}& {\displaystyle \frac{1}{\lambda }}=R({\displaystyle \frac{1}{{n_1}^2}}-{\displaystyle \frac{1}{{n_2}^2}})\\ & \Rightarrow {\displaystyle \frac{1}{\lambda }}=109679x({\displaystyle \frac{1}{4^2}}-{\displaystyle \frac{1}{6^2}})\\ & \Rightarrow \lambda =2.63x{10}^{-4}cm\end{array}$$